A black body is at a temperature of $$5760\,K.$$  The energy of radiation emitted by the body at wavelength $$250\,nm$$  is $${U_1},$$ at wavelength $$500\,nm$$  is $${U_2}$$ and that at $$1000\,nm$$  is $${U_3}.$$ Wien’s constant, $$b = 2.88 \times {10^6}nmK.$$    Which of the following is correct?

Solution :
Given, temperature, $${T_1} = 5760\,K$$
Since, it is given that energy of radiation emitted by the body at wavelength $$250\,nm$$  in $${U_1},$$ at wavelength $$500\,nm$$  is $${U_2}$$ and that at $$1000\,nm$$   is $${U_3}.$$
$$\because $$ According to Wien’s law, we get $${\lambda _m}T = b$$
where, $$b =$$  Wien’s constant $$ = 2.88 \times {10^6}nmK$$
$$\eqalign{ & \Rightarrow {\lambda _m} = \frac{b}{T} \cr & \Rightarrow {\lambda _m} = \frac{{2.88 \times {{10}^6}nmK}}{{5760\,K}} \cr & \Rightarrow {\lambda _m} = 500\,nm \cr} $$
$$\therefore {\lambda _m} = $$    wavelength corresponding to maximum energy, so, $${U_2} > {U_1}.$$